`0.5` g of metal nitrate gave `0.43` g of metal sulphate

Options

A. The equivalent wt. of the metal is `0.38`
B. The atomic weight of the metal is 76
C. The atomic wt. of metal is `38 x ` (x = valency of metal )
D. The atomic weight of metal is 19

Answer

A, C

Explanation

Let metal nitrate = `M(NO_(3))_(x)`
Metal sulphate ` = M_(2)(SO_(4))_(x)`
Eq. of `Mn(NO_(3))_(x) = " Eq. of " M_(2)(SO_(4))_(x)`
Let Eqv. Wt. of metal = `E_(M)`
`(0.5)/(E_(M) + 62) = (0.43)/(E_(M) + 96/2) `
` rArr 0.5 E_(M) + 0.5 xx 48 = 0.43 xx E_(M) + 62 xx 0.43`
` rArr 0.5 E_(M) + 24 = 0.43 E_(M) + 26.66`
` rArr (0.5 - 0.43) E_(M) = 26.66 - 24`
` rArr 0.07 E_(M) = 2.66`
` :. E_(M) = (2.66)/(0.07) = 38`
As valence of metal is x
Atomic wt. = `38 x`